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# Sum of Functions

Let $$f(x)=x-\dfrac{1}{x}$$.

(a) If $$f(x)+f(y)=0$$, then prove or disprove that $$f(x^n)+f(y^n)=0$$ for all integers $$n$$.

(b) If $$f(x)+f(y)+f(z)=0$$, then prove or disprove that $$f(x^n)+f(y^n)+f(z^n)=0$$ for all integers $$n$$.

(c) Do the above two problems with the added restriction that $$x,y,z$$ are positive.

Note by Daniel Liu
2 years, 3 months ago

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(a) Counterexample: $$y=-x$$ and $$n=2a$$ for some positive integer $$a$$. $$f(x)+f(-x)=x-\dfrac{1}{x}-x+\dfrac{1}{x}=0$$, but $$f(x^n)+f((-x)^n)=f(x^{2a})+f(x^{2a})=x^{2a}-\dfrac{1}{x^{2a}}+x^{2a}-\dfrac{1}{x^{2a}}\neq0$$.

(b) Again, the counterexample is $$y=-x$$, $$z=1$$, and $$n=2a$$ for some positive integer $$a$$. $$f(x)+f(-x)+f(1)=x-\frac{1}{x}-x+\frac{1}{x}+1-1=0$$, but $$f(x^n)+f((-x)^n)+f(1^n)=f(x^{2a})+f(x^{2a})=f(1)=x^{2a}-\dfrac{1}{x^{2a}}+x^{2a}-\dfrac{1}{x^{2a}}+1-1\neq0$$.

(c) (part 1): Since $$x,y$$ are positive and $$f(x)+f(y)=0$$, $$x-\dfrac{1}{x}+y-\dfrac{1}{y}=0$$, and $$x+y=\dfrac{1}{x}+\dfrac{1}{y}$$. Adding the fractions on the RHS gives $$x+y=\dfrac{x+y}{xy}$$ so $$1=\dfrac{1}{xy}$$ or $$xy=1$$. Therefore, $$f(x^n)+f(y^n)=x^n-\dfrac{1}{x^n}+y^n-\dfrac{1}{y^n}=x^n-\dfrac{1}{x^n}+\dfrac{1}{x^n}-x^n=0$$. · 2 years, 3 months ago

Sorry, your answer for part (a) is wrong. Also, the restriction of part (c) actually matters.

EDIT: thanks, fixed! · 2 years, 3 months ago